Missing Ranges

Question:

Given a sorted integer array where the range of elements are [0, 99] inclusive, return its missing ranges.

For example, given [0, 1, 3, 50, 75], return [“2”, “4->49”, “51->74”, “76->99”]

Example Questions Candidate Might Ask:

Q: What if the given array is empty?

A: Then you should return [“0->99”] as those ranges are missing.

Q: What if the given array contains all elements from the ranges?

A: Return an empty list, which means no range is missing.

Solution:

Compare the gap between two neighbor elements and output its range, simple as that

right? This seems deceptively easy, except there are multiple edge cases to consider, such

as the first and last element, which does not have previous and next element respectively.

Also, what happens when the given array is empty? We should output the range “0->99”.

As it turns out, if we could add two “artificial” elements, –1 before the first element and

100 after the last element, we could avoid all the above pesky cases.

Further Thoughts:

i. List out test cases.

ii. You should be able to extend the above cases not only for the range [0,99],

but any arbitrary range [start, end].

public List<String> findMissingRanges(int[] vals, int start, int end) {
List<String> ranges = new ArrayList<>();
int prev = start - 1;
for (int i = 0; i <= vals.length; i++) {
int curr = (i == vals.length) ? end + 1 : vals[i];
if (curr - prev >= 2) {
ranges.add(getRange(prev + 1, curr - 1));
}
prev = curr;
}
return ranges;
}
private String getRange(int from, int to) {
return (from == to) ? String.valueOf(from) : from + "->" + to;
}